We need to solve the inequality $\frac{x}{a} + \frac{1-3x}{2} < \frac{x+2}{4a}$, given that $a \neq 0$.

1. Problem Description

We need to solve the inequality $\frac{x}{a} + \frac{1-3x}{2} < \frac{x+2}{4a}$, given that $a \neq 0$.

2. Solution Steps

First, we multiply both sides of the inequality by $4a$ to eliminate the fractions. We need to consider two cases: $a>0$ and $a<0$.

Case 1: $a > 0$

Since $a > 0$, multiplying by $4a$ will not change the direction of the inequality.

$4a(\frac{x}{a} + \frac{1-3x}{2}) < 4a(\frac{x+2}{4a})$

$4x + 2a(1-3x) < x+2$

$4x + 2a - 6ax < x+2$

$3x - 6ax < 2 - 2a$

$x(3 - 6a) < 2 - 2a$

Now, we consider two subcases: $3-6a>0$ and $3-6a<0$.

Subcase 1.1: $3-6a>0$ which means $3>6a$, or $a<\frac{1}{2}$. Since we also have $a>0$, we have $0 < a < \frac{1}{2}$.

In this subcase, we divide both sides by $3-6a$ without changing the inequality sign:

$x < \frac{2-2a}{3-6a} = \frac{2(1-a)}{3(1-2a)}$

Subcase 1.2: $3-6a<0$ which means $3<6a$, or $a>\frac{1}{2}$.

In this subcase, we divide both sides by $3-6a$ and change the inequality sign:

$x > \frac{2-2a}{3-6a} = \frac{2(1-a)}{3(1-2a)}$

Case 2: $a < 0$

Since $a < 0$, multiplying by $4a$ will change the direction of the inequality.

$4a(\frac{x}{a} + \frac{1-3x}{2}) > 4a(\frac{x+2}{4a})$

$4x + 2a(1-3x) > x+2$

$4x + 2a - 6ax > x+2$

$3x - 6ax > 2 - 2a$

$x(3 - 6a) > 2 - 2a$

Since $a<0$, then $-6a>0$, so $3-6a>3>0$.

Therefore, we divide by $3-6a$ without changing the inequality sign:

$x > \frac{2-2a}{3-6a} = \frac{2(1-a)}{3(1-2a)}$

In summary:

If $0 < a < \frac{1}{2}$, then $x < \frac{2(1-a)}{3(1-2a)}$.

If $a > \frac{1}{2}$, then $x > \frac{2(1-a)}{3(1-2a)}$.

If $a < 0$, then $x > \frac{2(1-a)}{3(1-2a)}$.

3. Final Answer

If $0 < a < \frac{1}{2}$, then $x < \frac{2(1-a)}{3(1-2a)}$.

If $a > \frac{1}{2}$, then $x > \frac{2(1-a)}{3(1-2a)}$.

If $a < 0$, then $x > \frac{2(1-a)}{3(1-2a)}$.